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GUJCET · Physics · Current Electricity
Two wires of same material having lengths and radius in the ratio of \(3: 4\) and \(3: 2\) respectively are connected in parallel with a potential source of 6 V . The ratio of currents flowing through them \(I_1: I_2=\) ________.
- A \(1: 3\)
- B \(3: 1\)
- C \(1: 2\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(B) \(3: 1\)
Step-by-step Solution
Detailed explanation
(B)
Both wires are connected in parallel.
\(I_1 R_1=I_2 R_2\)
\(\therefore \frac{ I _1}{ I _2}=\frac{ R _2}{ R _1}=\frac{\rho \frac{l_2}{\pi r_2^2}}{\rho \frac{l_1}{\pi r_1^2}}\)
\(\therefore \frac{ I _1}{ I _2}=\frac{l_2}{l_1} \times\left(\frac{r_1}{r_2}\right)^2\)
\(\begin{array}{l}\therefore \frac{I_1}{I_2}=\frac{4}{3} \times\left(\frac{3}{2}\right)^2 \\ \therefore \frac{I_1}{I_2}=\frac{3}{1}\end{array}\)
Both wires are connected in parallel.
\(I_1 R_1=I_2 R_2\)
\(\therefore \frac{ I _1}{ I _2}=\frac{ R _2}{ R _1}=\frac{\rho \frac{l_2}{\pi r_2^2}}{\rho \frac{l_1}{\pi r_1^2}}\)
\(\therefore \frac{ I _1}{ I _2}=\frac{l_2}{l_1} \times\left(\frac{r_1}{r_2}\right)^2\)
\(\begin{array}{l}\therefore \frac{I_1}{I_2}=\frac{4}{3} \times\left(\frac{3}{2}\right)^2 \\ \therefore \frac{I_1}{I_2}=\frac{3}{1}\end{array}\)
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